Let A, B, C be three points in $x y$-plane, whose position vector are given by $\sqrt{3} \hat{i}+\hat{j},…
Let A, B, C be three points in $x y$-plane, whose position vector are given by $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $\mathrm{a} \hat{i}+(1-\mathrm{a}) \hat{j}$ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ is $\frac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is :
$2$
$9 / 2$
$1$
$0$
Solution
Equation of line in the internal bisector of $O \dot{A}$ and $O B$ is $(\sqrt{3}+1) \hat{i}+(\sqrt{3}+1) \hat{j}$ $\Rightarrow$ line will be $y=x \Rightarrow x-y=0$ $D=\left|\frac{a-(1-a)}{\sqrt{a^2+(1-a)^2}}\right|=\frac{9}{\sqrt{2}}$ $\begin{aligned} & (2 a-1)^2=\frac{81}{2}\left(a^2+(1-a)^2\right) \\ & \Rightarrow 2\left(4 a^2-4 a+1\right)=81 a^2+81 a^2-162 a-81 \\ & \Rightarrow 162 a^2-162 a+81-8 a^2+8 a-2=0 \\ & \Rightarrow 154 a^2-154 a+79=0 \\ & \text { Sum of values }=-\frac{(-154)}{154}=1\end{aligned}$
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